Question

Р(x, y), Q(–2, –3), and R(2, 3) are the vertices of a right triangle PQR right-angled at P. Find the relationship between x and y. Hence, find all possible values of x for which y = 2.

Answer 1

By using the distance formula,

PQ = `sqrt((x + 2)^2 + (y + 3)^2)`

QR = `sqrt((-2 - 2)^2 + (- 3 - 3)^2)`

= `sqrt((-4)^2 + (-6)^2)`

= `sqrt(16 + 36)`

= `sqrt52`

PR = `sqrt((x - 2)^2 + (y - 3)^2)`

By the Pythagorean theorem,

RQ² = RP² + PQ²

`(sqrt52)^2 = (sqrt((x - 2)^2 + (y - 3)^2))^2 + (sqrt((x + 2)^2 + (y + 3)^2))^2`

52 = x² + 4 − 4x + y² + 9 − 6y + x² + 4 + 4x + y² + 9 + 6y

52 = 2x²+ 2y² + 26

52 − 26 = 2x² + 2y²

2(x² + y²) = 26

x² + y² = `26/2`

x² + y² = 13

Given y = 2

x² + (2)² = 13

x² + 4 = 13

x² = 13 − 4

x² = 9

x = ± 3